14.5 problem Problem 31

Internal problem ID [2871]
Internal file name [OUTPUT/2363_Sunday_June_05_2022_03_00_53_AM_61815175/index.tex]

Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.7. page 704
Problem number: Problem 31.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "linear", "first_order_ode_lie_symmetry_lookup"

Maple gives the following as the ode type

[[_linear, `class A`]]

\[ \boxed {y^{\prime }+3 y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

14.5.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=3\\ q(t) &=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right . \end {align*}

Hence the ode is \begin {align*} y^{\prime }+3 y = \left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right . \end {align*}

The domain of \(p(t)=3\) is \[ \{-\infty

14.5.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )+3 Y \left (s \right ) = \frac {1-{\mathrm e}^{-s}}{s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-1+3 Y \left (s \right ) = \frac {1-{\mathrm e}^{-s}}{s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = -\frac {-1+{\mathrm e}^{-s}-s}{s \left (s +3\right )} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (-\frac {-1+{\mathrm e}^{-s}-s}{s \left (s +3\right )}\right )\\ &= \frac {\operatorname {Heaviside}\left (1-t \right )}{3}+\frac {\operatorname {Heaviside}\left (t -1\right ) {\mathrm e}^{-3 t +3}}{3}+\frac {2 \,{\mathrm e}^{-3 t}}{3} \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {1}{3} & t <1 \\ \frac {2 \,{\mathrm e}^{-3}}{3}+\frac {2}{3} & t =1 \\ \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {{\mathrm e}^{-3 t +3}}{3} & 1

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {1}{3} & t <1 \\ \frac {2 \,{\mathrm e}^{-3}}{3}+\frac {2}{3} & t &=1 \\ \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {{\mathrm e}^{-3 t +3}}{3} & 1

Verification of solutions

\[ y = \left \{\begin {array}{cc} \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {1}{3} & t <1 \\ \frac {2 \,{\mathrm e}^{-3}}{3}+\frac {2}{3} & t =1 \\ \frac {2 \,{\mathrm e}^{-3 t}}{3}+\frac {{\mathrm e}^{-3 t +3}}{3} & 1

14.5.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }+3 y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right ., y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-3 y+\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }+3 y=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right . \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+3 y\right )=\mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }+3 y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=3 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{3 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{3 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{3 t} \left (\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \end {array}\right .\right )d t +c_{1}}{{\mathrm e}^{3 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\left \{\begin {array}{cc} 0 & t \le 0 \\ \frac {{\mathrm e}^{3 t}}{3}-\frac {1}{3} & t \le 1 \\ \frac {{\mathrm e}^{3}}{3}-\frac {1}{3} & 1

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`
 

✓ Solution by Maple

Time used: 3.688 (sec). Leaf size: 43

dsolve([diff(y(t),t)+3*y(t)=piecewise(0<=t and t<1,1,t>=1,0),y(0) = 1],y(t), singsol=all)
 

\[ y \left (t \right ) = \frac {\left (\left \{\begin {array}{cc} 1+2 \,{\mathrm e}^{-3 t} & t <1 \\ 2 \,{\mathrm e}^{-3}+2 & t =1 \\ 2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3} & 1

✓ Solution by Mathematica

Time used: 0.069 (sec). Leaf size: 47

DSolve[{y'[t]+3*y[t]==Piecewise[{{1,0<=t<1},{0,t >= 1}}],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-3 t} & t\leq 0 \\ \frac {1}{3} e^{-3 t} \left (2+e^3\right ) & t>1 \\ \frac {1}{3}+\frac {2 e^{-3 t}}{3} & \text {True} \\ \end {array} \\ \end {array} \]